1S11: Calculus for students in Science
Dr. Vladimir Dotsenko
TCD
Lecture 25
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 25
1/1
Antiderivatives: reminder
It is often important for applications to solve the reverse problem: assume
that we know the derivative of a function, what can we say about the
original function? For example, we see the instantaneous velocity on the
speedometer all the time, and want to know how far we progressed over
the given period of time.
Definition. A function F is called an antiderivative of the given
function f if
dF (x)
= f (x).
dx
Suppose that F1 and F2 are two antiderivatives of the same function f .
Then
dF1 (x) dF2 (x)
d(F1 (x) − F2 (x))
=
−
= f (x) − f (x) = 0,
dx
dx
dx
which implies that F1 (x) − F2 (x) is a constant c. Vice versa, if F is an
antiderivative of f , then G (x) = F (x) + c is also an antiderivative of f .
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 25
2/1
Integral calculus: antiderivatives
There is an alternative “integral” notation for antiderivatives:
Z
f (x) dx = F (x) + C
R
denotes the fact that F is an antiderivative of f . The expression f (x) dx
on the left side of that equation is called the indefinite integral.
Examples. Some examples of indefinite integrals are shown below:
derivative
formula
(x 2 )′ = 2x
(sin x)′ = cos x
(tan x) = cos12 x
(ln x)′ = 1
n+1 ′ x
x
= xn
n+1
Dr. Vladimir Dotsenko (TCD)
equivalent
integration
formula
R
2
R 2x dx = x + C
cos x dx = sin x + C
R dx
2 x = tan x + C
Rcosdx
x = ln x + C
R n
n+1
x dx = xn+1 + C
1S11: Calculus for students in Science
Lecture 25
3/1
Rules for antiderivatives
Theorem. Suppose that F (x) and G (x) are antiderivatives of f (x) and
g (x) respectively, and that c is a constant. Then
R
cf (x) dx = cF (x) + C ,
R
[f (x) + g (x)] dx = F (x) + G (x) + C ,
R
[f (x) − g (x)] dx = F (x) − G (x) + C .
Proof. Just use rules for differentiation. For example,
(F (x) − G (x))′ = F ′ (x) − G ′ (x) = f (x) − g (x).
Another way to write the same is
R
R
cf (x) dx = c f (x) dx,
R
R
R
[f (x) + g (x)] dx = f (x) dx + g (x) dx,
R
R
R
[f (x) − g (x)] dx = f (x) dx − g (x) dx.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 25
4/1
Rules for antiderivatives
These rules should be handled with care, leaving the arbitrary constant till
the very end of the computation, so that we do not write anything like that
2
Z
Z
x
2
x + C = 2x dx = 2 x dx = 2
+ C = x 2 + 2C ,
2
concluding then that C = 2C , so C = 0.
Of course, there is a more general rule which combines addition and
multiplication by constants:
Z
[c1 f1 (x) + c2 f2 (x) + · · · + cn fn (x)] dx =
Z
Z
Z
= c1 f1 (x) dx + c2 f2 (x) dx + · · · + cn fn (x) dx.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 25
5/1
Rules for antiderivatives
Theorem. (u-substitution) Suppose that F (x) is an antiderivative of
f (x). Then
Z
[f (g (x))g ′ (x)] dx = F (g (x)) + C .
Proof. Applying the chain rule, we obtain
(F (g (x)))′ = F ′ (g (x))g ′ (x) = f (g (x))g ′ (x),
as required.
Example. Let us evaluate the integral
Z
(x 2 + 1)50 · 2x dx.
Let g (x) = x 2 + 1. Noticing that (x 2 + 1)′ = 2x, we can write the
integral as
Z
Z
(x 2 + 1)51
g (x)51
2
50
+C =
+ C.
(x + 1) · 2x dx = g (x)50 g ′ (x) dx =
51
51
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 25
6/1
u-substitution: more examples
Example. Let us evaluate the integral
Z
cos x
dx.
sin2 x
Denoting g (x) = sin x, we get
Z
Z
cos x
1
1
1
+C =−
+ C.
g ′ (x) dx = −
dx =
2
2
g (x)
g (x)
sin x
sin x
Example. Let us evaluate the integral
√
Z
cos x
√
dx.
x
√
Denoting g (x) = x, we get
√
Z
Z
√
cos x
√
dx = cos g (x) · 2g ′ (x) dx = 2 sin g (x) + C = 2 sin x + C .
x
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 25
7/1
Rules for antiderivatives
We already know that differentiating products is not as easy as
differentiating sums. Same is true for antiderivatives.
Theorem. (Integration by parts) Suppose that F (x) and G (x) are
antiderivatives of f (x) and g (x) respectively. Then
Z
Z
[f (x)G (x)] dx + [F (x)g (x)] dx = F (x)G (x) + C .
Proof. Applying the product rule, we obtain
(F (x)G (x))′ = F ′ (x)G (x) + F (x)G ′ (x) = f (x)G (x) + F (x)g (x),
as required.
This rule is usually applied in the following way:
Z
Z
[f (x)G (x)] dx = F (x)G (x) − [F (x)g (x)] dx,
meaning that computing the antiderivative of f (x)G (x) can be reduced to
computing the antiderivative of F (x)g (x), and the latter one can be
substantially simpler to compute.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 25
8/1
Integration by parts: examples
R
Example. Let us evaluate xe x dx. Using integration by parts with
G (x) = x, f (x) = e x (in which case g (x) = 1, and we can take
F (x) = e x ), we get
Z
Z
x
x
xe dx = xe − e x · 1 dx = xe x − e x + C .
Note that if we instead apply integration by parts with f (x) = x,
2
G (x) = e x (so that F (x) = x2 , g (x) = e x ), we get
Z
x2
xe dx = e x −
2
x
Z
x2 x
e dx,
2
and our computation expressed the given integral using what is a more
complicated indefinite integral!
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 25
9/1
Integration by parts: examples
R
Example. Let us evaluate x 2 e x dx. Using integration by parts with
G (x) = x 2 , f (x) = e x (in which case g (x) = 2x, and we can take
F (x) = e x ), we get
Z
Z
2 x
2 x
x e dx = x e − 2xe x dx = x 2 e x − 2xe x + 2e x + C ,
R
since we already know xeRx dx.
Example. Let us evaluate ln x dx. Using integration by parts with
G (x) = ln x, f (x) = 1 (in which case g (x) = x1 , and we can take
F (x) = x), we get
Z
Z
1
ln x dx = x ln x − x · dx = x ln x − x + C .
x
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 25
10 / 1
Integration by parts: examples
R
Example. Let us evaluate e x sin x dx. Using integration by parts with
G (x) = sin x, f (x) = e x (in which case g (x) = cos x, and we can take
F (x) = e x ), we get
Z
Z
x
x
e sin x dx = e sin x − e x cos x dx.
Now we use integration by parts with G (x) = cos x, f (x) = e x (in which
case g (x) = − sin x, and we can take F (x) = e x ), so that we get
Z
Z
x
x
e cos x dx = e cos x + e x sin x dx,
so that
R
R
e x sin x dx = e x sin x − e x cos x − e x sin x dx, and
Z
1
e x sin x dx = (e x sin x − e x cos x) + C .
2
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 25
11 / 1